Question: $ \left(\dfrac{9}{64}\right)^{-\frac{1}{2}}$
Solution: $= \left(\dfrac{64}{9}\right)^{\frac{1}{2}}$ Figure out what goes in the blank: $\Big(? \Big)^{2}=\dfrac{64}{9}$ Figure out what goes in the blank: $\Big({\dfrac{8}{3}}\Big)^{2}=\dfrac{64}{9}$ So $\left(\dfrac{9}{64}\right)^{-\frac{1}{2}}=\left(\dfrac{64}{9}\right)^{\frac{1}{2}}=\dfrac{8}{3}$